calculate the image position and magnification for an object 3cm tall laced in front of mirror with a focal length of 12cm, if the object is placed at 2f... I got:

di: 24cm
and
m=-1

please tell me if this is right.. thank you so much :)

omg i had the same question jason says your right he found the answer in the text book

heyy that's right its in the physics book page 337 the answer is in the back of the book lol good job i need to learn how to do it lol :P

To calculate the image position and magnification of an object placed in front of a mirror, you can use the mirror equation and magnification formula.

The mirror equation is:
1/f = 1/do + 1/di

Where:
f = focal length of the mirror
do = object distance (distance of the object from the mirror)
di = image distance (distance of the image from the mirror)

The magnification formula is:
m = -di/do

Where:
m = magnification
di = image distance
do = object distance

Given:
focal length (f) = 12 cm
object distance (do) = 2f (since the object is placed at 2 times the focal length)

Using the mirror equation, we can solve for di:
1/12 = 1/(2f) + 1/di

1/12 = 1/24 + 1/di

Multiply both sides by 24di:
2di = 1 + 24

2di = 25

di = 25/2
di = 12.5 cm

Now, let's calculate the magnification (m):
m = -di/do

Since do = 2f = 2(12) = 24 cm:
m = -12.5/24
m ≈ -0.52

So, the image position (di) is approximately 12.5 cm and the magnification (m) is approximately -0.52.

Therefore, your calculated values for di and m are correct.